# Molar solubility

Molar solubility is the number of moles of a substance (the solute) that can be dissolved per liter of solution before the solution becomes saturated. It can be calculated from a substance's solubility product constant (Ksp) and stoichiometry. The units are mol/L, sometimes written as M.

## Derivation

Given excess of a simple salt AxBy in an aqueous solution of no common ions (A or B already present in the solution), the amount of it which enters solution can be calculated as follows:

The chemical equation for this salt would be:

${\ce {{{A}_{\mathit {x}}{B}_{\mathit {y}}}{(s)}->{\mathit {x}}{A}{(aq)}+{\mathit {y}}{B}{(aq)}}}$ where A, B are ions and x, y are coefficients...

1. The relationship of the changes in amount (of which mole is a unit), represented as N(∆), between the species is given by stoichiometry as follows:
$-{\frac {N_{{\ce {A}}_{x}{\ce {B}}_{y}(\Delta )}}{1}}={\frac {N_{{\ce {A(\Delta)}}}}{x}}={\frac {N_{{\ce {B(\Delta)}}}}{y}}\,$ which, when rearranged for ∆A and ∆B yields:

$N_{{\ce {A(\Delta)}}}=-xN_{{\ce {A}}_{x}{\ce {B}}_{y}(\Delta )}\,$ $N_{{\ce {B(\Delta)}}}=-yN_{{\ce {A}}_{x}{\ce {B}}_{y}(\Delta )}\,$ 2. The Deltas(∆), Initials(i) and Finals(f) relate very simply since, in this case, molar solubility is defined assuming no common ions are already present in the solution.
$N_{(i)}+N_{(\Delta )}=N_{(f)}\,$ Difference law
$N_{{{\ce {A}}}(i)}=0\,$ $N_{{{\ce {B}}}(i)}=0\,$ Which condense to the identities

$N_{{{\ce {A}}}(f)}=N_{{{\ce {A}}}(\Delta )}\,$ $N_{{{\ce {B}}}(f)}=N_{{{\ce {B}}}(\Delta )}\,$ 3. In these variables (with V for volume), the molar solubility would be written as:
$S_{0}=-{\frac {N_{{{\ce {A}}}_{x}{{\ce {B}}}_{y}(\Delta )}}{V}}\,$ 4. The solubility product expression is defined as:
$K_{sp}=[{{\ce {A}}}]^{x}[{{\ce {B}}}]^{y}\,$ These four sets of equations are enough to solve for S0 algebraically:

{\begin{aligned}K_{sp}&={\left({\frac {N_{{{\ce {A}}}(f)}}{V}}\right)}^{x}{\left({\frac {N_{{{\ce {B}}}(f)}}{V}}\right)}^{y}\\&={\left({\frac {N_{{\ce {A(\Delta )}}}}{V}}\right)}^{x}{\left({\frac {N_{{\ce {B(\Delta )}}}}{V}}\right)}^{y}\\&={\frac {{(N_{{\ce {A(\Delta )}}})}^{x}{(N_{{\ce {B(\Delta )}}})}^{y}}{V^{(x+y)}}}\\&={\frac {{(-xN_{{{\ce {A}}}_{x}{{\ce {B}}}_{y}(\Delta )})}^{x}{(-yN_{{{\ce {A}}}_{x}{{\ce {B}}}_{y}(\Delta )})}^{y}}{V^{(x+y)}}}\\&={\frac {{(-1)}^{x}{(x)}^{x}{(N_{{{\ce {A}}}_{x}{{\ce {B}}}_{y}(\Delta )})}^{x}{(-1)}^{y}{(y)}^{y}{(N_{{{\ce {A}}}_{x}{{\ce {B}}}_{y}(\Delta )})}^{y}}{V^{(x+y)}}}\\&=x^{x}y^{y}{\frac {{(-1)}^{x}{(-1)}^{y}{(N_{{{\ce {A}}}_{x}{{\ce {B}}}_{y}(\Delta )})}^{x}{(N_{{{\ce {A}}}_{x}{{\ce {B}}}_{y}(\Delta )})}^{y}}{V^{(x+y)}}}\\&=x^{x}y^{y}{\frac {{(-1)}^{(x+y)}{(N_{{{\ce {A}}}_{x}{{\ce {B}}}_{y}(\Delta )})}^{(x+y)}}{V^{(x+y)}}}\end{aligned}} {\begin{aligned}{\frac {K_{sp}}{x^{x}y^{y}}}&={\left(-{\frac {N_{{{\ce {A}}}_{x}{{\ce {B}}}_{y}(\Delta )}}{V}}\right)}^{(x+y)}\\&={\left(S_{0}\right)}^{(x+y)}\end{aligned}} Hence;

$S_{0}={\sqrt[{(x+y)}]{\frac {K_{sp}}{x^{x}y^{y}}}}$ ## Simple calculation

If the solubility product constant (Ksp) and dissociation product ions are known, the molar solubility can be computed without the aforementioned equation.

### Example

Ionic substance AB2 dissociates into A and 2B, or one mole of ion A and two moles of ion B. The soluble ion dissociation equation can thus be written as:

${\ce {AB2(s) -> A{(aq)}+ 2B{(aq)}}}$ where the corresponding solubility product equation is:

$K_{sp}={\ce {[A][B]^2}}$ If the final concentration of A is x, then that of B must be 2x. Inserting these final concentrations into the solubility product equation gives:

{\begin{aligned}K_{sp}&=(x)(2x)^{2}\\&=(x)(4x^{2})\\&=4x^{3}\end{aligned}} If Ksp is known, x can be computed, which is the molar solubility.